# Homework 4

*Due by 9:00pm on Wednesday, 2/27/2019*

## Instructions

Download hw04.zip. Inside the archive, you will find starter files for the questions in this homework, along with a copy of the OK autograder.

**Submission:** When you are done, submit with `python3 ok --submit`

. You may submit more than once before the deadline; only the final submission will be scored. Check that you have successfully submitted your code on okpy.org. See this article for more instructions on okpy and submitting assignments.

**Readings:** This homework relies on following references:

## Required questions

### Question 1: arange

Implement the function `arange`

, which behaves just like np.arange(start, end, step) from Data 8. You only need to support positive values for step.

```
def arange(start, end, step=1):
"""
arange behaves just like np.arange(start, end, step).
You only need to support positive values for step.
>>> arange(1, 3)
[1, 2]
>>> arange(0, 25, 2)
[0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24]
>>> arange(999, 1231, 34)
[999, 1033, 1067, 1101, 1135, 1169, 1203]
"""
"*** YOUR CODE HERE ***"
```

Use OK to test your code:

`python3 ok -q arange`

### Question 2: Repeated

Implement `repeated(f, n)`

:

`f`

is a one-argument function that takes a number and returns another number.`n`

is a positive integer

`repeated`

returns another function that, when given an argument `x`

, will
compute `f(f(....(f(x))....))`

(apply `f`

a total `n`

times). For example,
`repeated(square, 3)(42)`

evaluates to `square(square(square(42)))`

.

```
def repeated(f, n):
"""Return the function that computes the nth application of f.
>>> def increment(x):
... return x + 1
>>> def square(x):
... return x**2
>>> def triple(x):
... return x*3
>>> add_three = repeated(increment, 3)
>>> add_three(5)
8
>>> repeated(triple, 5)(1) # 3 * 3 * 3 * 3 * 3 * 1
243
>>> repeated(square, 2)(5) # square(square(5))
625
>>> repeated(square, 4)(5) # square(square(square(square(5))))
152587890625
"""
"*** YOUR CODE HERE ***"
```

*Hint*: You may find it convenient to use `compose1`

from the textbook:

```
def compose1(f, g):
"""Return a function h, such that h(x) = f(g(x))."""
def h(x):
return f(g(x))
return h
```

Use OK to test your code:

`python3 ok -q repeated`

### Question 3: Double

Using `repeated`

define a function `double`

that takes a function of
one argument as an argument and returns a function that applies the
original function twice. For example, if `inc`

is a function that
returns `1`

more than its argument, then `double(inc)`

should be a
function that returns two more:

```
def double(f):
""" Return a function that applies f twice.
f -- a function that takes one argument
>>> def square(x):
... return x**2
>>> double(square)(2)
16
"""
"*** YOUR CODE HERE ***"
```

Use OK to test your code:

`python3 ok -q double`

### Question 4: Count van Count

Consider the following implementations of `count_factors`

and `count_primes`

:

```
def count_factors(n):
"""Return the number of positive factors that n has."""
i, count = 1, 0
while i <= n:
if n % i == 0:
count += 1
i += 1
return count
def count_primes(n):
"""Return the number of prime numbers up to and including n."""
i, count = 1, 0
while i <= n:
if is_prime(i):
count += 1
i += 1
return count
def is_prime(n):
return count_factors(n) == 2 # only factors are 1 and n
```

The implementations look quite similar! Generalize this logic by writing a
function `count_cond`

, which takes in a two-argument predicate function ```
condition(n,
i)
```

. `count_cond`

returns a count of all the numbers from 1 to `n`

that satisfy
`condition`

.

Note: A predicate function is a function that returns a boolean (`True`

or `False`

).

```
def count_cond(condition, n):
"""
>>> def divisible(n, i):
... return n % i == 0
>>> count_cond(divisible, 2) # 1, 2
2
>>> count_cond(divisible, 4) # 1, 2, 4
3
>>> count_cond(divisible, 12) # 1, 2, 3, 4, 6, 12
6
>>> def is_prime(n, i):
... return count_cond(divisible, i) == 2
>>> count_cond(is_prime, 2) # 2
1
>>> count_cond(is_prime, 3) # 2, 3
2
>>> count_cond(is_prime, 4) # 2, 3
2
>>> count_cond(is_prime, 5) # 2, 3, 5
3
>>> count_cond(is_prime, 20) # 2, 3, 5, 7, 11, 13, 17, 19
8
"""
"*** YOUR CODE HERE ***"
```

Use OK to test your code:

`python3 ok -q count_cond`

### Question 5: Match and Apply

Sometimes when we are given a dataset, we need to alter it for specific values. For example, say we have a table with one column being people's names and the other being the price they have to pay.

We can use a list of pairs for this:

`[["Jessica", 5], ["Andrew", 9], ["Alex", 2], ["Amir", 11], ["John", 3], ["Lyric", 2]]`

The first value in each pair is the name, the second is the price.

Now, let's say we want to give a discount to specific people. We have a discount function that we want to apply to the person's price. Now, we need a function that will only apply the discount function to specific people.

Implement `match_and_apply(pairs, function)`

:

`pairs`

is a list of pairs.`function`

is some function

`match_and_apply`

returns a function such that when the function is given an input that
matches the first of a pair, returns the result of applying `function`

to the second value in the pair.

```
def match_and_apply(pairs, function):
"""
>>> pairs = [[1, 2], [3, 4], [5, 6], [7, 8], [9, 0]]
>>> def square(num):
... return num**2
>>> func = match_and_apply(pairs, square)
>>> result = func(3)
>>> result
16
>>> result = func(1)
>>> result
4
>>> result = func(7)
>>> result
64
>>> result = func(15)
>>> print(result)
None
"""
"*** YOUR CODE HERE ***"
```

Use OK to test your code:

`python3 ok -q match_and_apply`